Question

The state of hybridisation of boron and oxygen atoms in boric acid(${{H}_{3}}B{{O}_{3}}$ ) respectively are:
(a) $s{{p}^{2}}$ and $s{{p}^{2}}$
(b) $s{{p}^{2}}$ and $s{{p}^{3}}$
(c) $s{{p}^{3}}$ and $s{{p}^{2}}$
(d) $s{{p}^{3}}$ and $s{{p}^{3}}$

Answer 1

Hint: Process of combining two or more atomic orbitals from the same atom to form an entirely new different orbital from its components is called hybridisation. During the formation of hydroboric acid, one s and multiple p orbitals of boron and oxygen undergo hybridisation.

Complete answer:
- Boric acid or orthoboric acid has a formula ${{H}_{3}}B{{O}_{3}}$ . The central boron atom forms a bond with three hydroxyl groups. There are three electrons in the outermost valence shell of Boron and all the three can form sigma bonds. There is no lone pair of electrons on Boron; all are engaged in bond formation with the OH- group. Thus it has a trigonal shape and is $s{{p}^{2}}$ hybridised.
- Each oxygen atom forms a sigma bond with hydrogen and boron. The outermost shell of oxygen contains 6 electrons. Out of 2 electrons (one form a sigma bond with boron and other with hydrogen) are engaged in bond formation. Thus, there are 2 lone pairs of electrons on each oxygen atom. So, the oxygen atoms are tetrahedral and for such an atom, the hybridisation is $s{{p}^{3}}$ .

The structure of boric acid is shown as:

Thus, the state of hybridisation of boron is $s{{p}^{2}}$ and of oxygen is $s{{p}^{3}}$ in hydroboric acid.

So, the correct answer is “Option C”.


Note:
Tetrahedral arrangement in an atom is only seen when the hybridisation of the atom is $s{{p}^{3}}$ . In a trigonal arranged atom, the hybridisation would be $s{{p}^{2}}$ . In hydroboric acid, the boron and the hydroxyl groups all are in the same plane.